hi Blaster0
***How did you get -2 from log6,25/log0,4***
To be perfectly honest I just put the whole thing straight into a calculator and it spat out -2
However, since I am on a roll here I will explain why it is so!
$$\frac{log6.25}{log0.4}=log_{0.4}6.25$$ I did this by applying the change of base law the opposite way around to usual.
Let
$$x=log_{0.4}6.25$$
Rearranging this we get
$$6.25=0.4^x\\
2.5^2=0.4^x\\
\left(\frac{5}{2}\right)^2=\left(\frac{2}{5}\right)^x\\
\left(\frac{2}{5}\right)^{-2}=\left(\frac{2}{5}\right)^x\\
x\:=\:-2$$
Hey Blaster0,
When you are happy with your answer can we have a thank you and some thumbs up please.
Speaking for myself I put a lot of effort into these answers!
(Don't mind me I am just trying to train people to be polite. I figure that the more people who do it then the more normal it becomes and hopefully the 'idea' will catch on. )
Melody And no chris you can not have your glasses back - I like them!
this should be the last one for the night - It is 1:30 am
$$0.4^{((6-5x)/(2+5x))}<6.25\\\\
log0.4^{((6-5x)/(2+5x))} \frac{6-5x}{2+5x}log0.4 \frac{6-5x}{2+5x}>\frac{log6.25}{log0.4}\\\\
\frac{6-5x}{2+5x}>\frac{log6.25}{log0.4}\\\\
\frac{6-5x}{2+5x}>-2\\\\
(2+5x)^2\frac{6-5x}{2+5x}>-2(2+5x)^2\\\\
(2+5x)(6-5x)>-2(25x^2+20x+4)\\\\
12-10x+30x-25x^2>-50x^2-40x-8\\\\$$
$$25x^2+60x+20>0\\\\
5x^2+12x+4>0\\\\
5x^2+10x+2x+4>0\\\\
5x(x+2)+2(x+2)>0\\\\
(5x+2)(x+2)>0\\\\
\mbox{concave up parabola roots are -2 and -2/5}\\\\
x<-2 \mbox{ OR }x>\frac{-2}{5}$$
Okay I decided to check this by plotting the original question. And to my delight it appears to be correct!
$${log}_{10}\left({\mathtt{6.25}}\right) = {\mathtt{0.795\: \!880\: \!017\: \!344\: \!075\: \!2}}$$
$${log}_{10}\left({\mathtt{0.4}}\right) = -{\mathtt{0.397\: \!940\: \!008\: \!672\: \!037\: \!6}}$$
If you divide the first by the second, the result is -2.
hi Blaster0
***How did you get -2 from log6,25/log0,4***
To be perfectly honest I just put the whole thing straight into a calculator and it spat out -2
However, since I am on a roll here I will explain why it is so!
$$\frac{log6.25}{log0.4}=log_{0.4}6.25$$ I did this by applying the change of base law the opposite way around to usual.
Let
$$x=log_{0.4}6.25$$
Rearranging this we get
$$6.25=0.4^x\\
2.5^2=0.4^x\\
\left(\frac{5}{2}\right)^2=\left(\frac{2}{5}\right)^x\\
\left(\frac{2}{5}\right)^{-2}=\left(\frac{2}{5}\right)^x\\
x\:=\:-2$$
Hey Blaster0,
When you are happy with your answer can we have a thank you and some thumbs up please.
Speaking for myself I put a lot of effort into these answers!
(Don't mind me I am just trying to train people to be polite. I figure that the more people who do it then the more normal it becomes and hopefully the 'idea' will catch on. )
Melody And no chris you can not have your glasses back - I like them!